###### Correlation

Covariance Covariance is a measure of how two variables move together. The sample... **Read More**

We can calculate and interpret probabilities of random variables that assume either the uniform distribution or the binomial distribution.

As you will recall**, **under the uniform distribution, all possible outcomes have equal probabilities. A good example might be the throw of a die, in which case each of the outcomes, 1, 2, 3, 4, 5, and 6, has a 1/6 probability of occurrence.

Define the random variable X such that X_{i} = i, where i = 1, 2, 3, …, k)

$$ P(X = x) = \cfrac {1}{k} $$

$$ \text{Mean} =\cfrac {(k + 1)}{2} $$

$$ \text{Variance} = \cfrac {(k^2 – 1)}{12} $$

A uniform distribution function has been defined as:

X = {1, 3, 5, 7, 9}

Calculate P(X = 5) and F(5).

**Solution**

$$ \begin{align*} P(X = x) & = \cfrac {1}{k} \\ K & = 5\\ \end{align*} $$

Therefore,

$$ P(X = 5) =\cfrac {1}{5} = 0.2 $$

F(5) is the cumulative probability of an outcome less than or equal to 5. Thus,

$$ \begin{align*} F(5) & = P(X = 1) + P(X = 3) + P(X = 5) \\ & = 0.2 + 0.2 + 0.2 \\ & = 0.6 \\ \end{align*} $$

The binomial distribution is a sequence of n Bernoulli trials where the outcome for every trial can be a success or a failure.

Suppose the probability of a success is θ:

$$ P\left( X=x \right) =\left( \begin{matrix} n \\ x \end{matrix} \right) { \theta }^{ x }\left( 1-\theta \right) ^{ n-x },x=0,1,2,…,n;0<\theta <1 $$

Where

$$ \left( \begin{matrix} n \\ x \end{matrix} \right) ={ \quad }^{ n }{ C }_{ x }=\frac { n! }{ \left( n-x \right) !x! } $$

$$ \text{Mean of X} = n \theta $$

$$ \text{Variance of X} = n \theta (1 – \theta )$$

Note: sometimes the probability of success can be denoted by “p” and (1-p) to denote the probability of failure.

The probability of surviving an attack by a certain disease is 60%. What is the probability that at least 11 out of a group of 12 people who catch the disease will survive?

**Solution**

The number of survivors is distributed binomially with parameters n = 12 and θ = 0.6:

$$ \begin{align*} P(X \ge 11) & = P(X = 11 \text{ or } 12) \\ &={ _{ 12 }{ C }_{ 11 }{ \ast 0.6 }^{ 11 }*{ 0.4 }^{ 1 }+_{ 12 }{ C }_{ 12 }{ \ast 0.6 }^{ 12 } } \\ &= \cfrac {12!}{(1!11!)} * 0.6^{11} * 0.4^1 + \cfrac {12!}{(0!12!)} *0.6^{12} \\ &= 12 * 0.003628 * 0.4 + 0.6^{12} \\ &= 0.01959 \end{align*} $$

QuestionA bowl contains blue and orange balls. The probability of drawing a blue ball in any attempt is 0.5. Suppose you draw 5 balls from the bag, what is the probability that 3 of the 5 balls drawn are blue?

A. 0.3125

B. 0.03125

C. 0.25

SolutionThe correct answer is A.

P(X = 3) =

_{5}C_{3 }* 0.5^{3}* 0.5^{2}= 10 * 0.03125

= 0.3125